Question 12 & 13, Exercise 3.3

Solutions of Question 12 & 13 of Exercise 3.3 of Unit 03: Vectors. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Prove that the angle in a semicircle is right angle.

We are considering a triangle inside a semicircle as shown. We have to show $\overrightarrow{B A} \cdot \overrightarrow{A C}=0$.

We see in figure that: $|\vec{a}|=\vec{b}|=| \vec{c} \mid=$ radius of circle.

Also $\vec{b}=-\vec{c}$ parallel but opposile in direction. From $\triangle A B O$, we have \begin{align}\overrightarrow{O B}+\overrightarrow{A B}&=\overrightarrow{O A}\\ \Rightarrow \overrightarrow{B A}&=\overrightarrow{O A}-\overrightarrow{O B}=\vec{a}-\vec{b}...(1)\end{align} Also from $\triangle A C O$, we have \begin{align}\overrightarrow{O A}+\overrightarrow{A C}&=\overrightarrow{O C}\\ \Rightarrow \overrightarrow{A C}&=\overrightarrow{O C}-\overrightarrow{O A}=\vec{c}-\vec{a} \text {...(2) }\end{align} Now \begin{align}\overrightarrow{B A} \cdot \overrightarrow{A C}& =(\vec{a}-\vec{b}) \cdot(\vec{c}-\vec{a}) \\ & =\vec{a} \cdot \vec{c}-\vec{a} \cdot \vec{a}-\vec{b} \cdot \vec{c}+\vec{b} \cdot \vec{a} \\ & =-\vec{a} \cdot \vec{b}-|\vec{a}|^2+|\vec{c}|^2+\vec{b} \cdot \vec{a}\quad by \quad \vec{a} \cdot \vec{a}=|\vec{a}|^2 and\quad \vec{b}=-\vec{c}\\ \Rightarrow \overrightarrow{B A} \cdot \overrightarrow{A C}&=0 \quad \because \quad|\vec{a}|=|\vec{c}|\end{align} Triangle formed looking in semicircle is right angle triangle.

Three or more than three lines intersecting at a single common point are known as concurrent lines and this property of intersecting at a single point is known as concurrency.

Prove that the perpendicular bisectors of the sides of a triangle are concurrent.

Let us considering a triangle $A B C$.
The position vectors of vertices $A, B$ and $C$ are $\vec{a}, \vec{b}$ and $\vec{c}$ respectively.
$D, E$ and $F$ are the midpoints of sides $B C$, $C A$ and $A B$ respectively.
Let the sides bisectors of $B C$ and $C A$ intersect at point $O$ as shown in figure.
We have to show that side bisector of $A B$ also passes through $O$.
Since $D E$ and $F$ are the midpoints of sides shown
\begin{align}\therefore \quad \overrightarrow{O D}&=\dfrac{\vec{b}+\vec{c}}{2}, \quad \overrightarrow{O E}=\dfrac{\vec{a}+\vec{c}}{2} \text { and } \\ \overrightarrow{O F}&=\dfrac{\vec{a}+\vec{b}}{2} . \\ &\text { Now } \overrightarrow{O D} \perp \overrightarrow{B C} \quad \\ \therefore \quad \overrightarrow{O D} \cdot \overrightarrow{B C}&=0 \\ \Rightarrow \dfrac{\vec{b}-\vec{c}}{2} \cdot(\vec{c}-\vec{b})&=0 \quad \because \overrightarrow{B C}=\vec{c}-\vec{b} . \\ \Rightarrow \quad \dfrac{|\vec{c}|^2-|\vec{b}|^2}{2}&=0 \ldots \ldots \ldots \ldots \ldots(1) \\ \text { Also } \overrightarrow{O E} \perp \overrightarrow{C A} \quad \therefore \quad \overrightarrow{O E} \cdot \overrightarrow{C A}&=0\end{align} \begin{align} \Rightarrow \dfrac{\vec{c}+\vec{a}}{2} \cdot(\vec{a}-\vec{c})&=0 \quad \because \overrightarrow{C A}=\vec{c}-\vec{a} \\ \Rightarrow \dfrac{|\vec{a}|^2-|\vec{c}|^2}{2}&=0 \ldots \ldots \ldots \ldots \ldots(2)\end{align} Adding (1) and (2), we get
\begin{align}\dfrac{|\vec{c}|^2-|\vec{b}|^2}{2}+\dfrac{|\vec{a}|^2-|\vec{c}|^2}{2}&=0 \\ \Rightarrow \dfrac{|\vec{a}|^2-|\vec{b}|^2}{2}&=0 \\ \Rightarrow \dfrac{(\vec{a}-\vec{b}) \cdot(\vec{a}+\vec{b})}{2}&=0 \\ \Rightarrow \dfrac{\vec{a}+\vec{b}}{2} \cdot(\vec{a}-\vec{b})&=0 \\ \Rightarrow \overrightarrow{O F} \perp \overrightarrow{A B}&=0\end{align} $\Rightarrow$ sid: bisector of $A B$ also passes through $O$, thus the three side bisectors are concurrent.